Titanium occurs in the magnetic mineral ilmenite (FeTiO₃), which is often found mixed with sand. The ilmenite can be separated from the sand with magnets. The titanium can then be extracted from the ilmenite by the following set of reactions: FeTiO₃(s) + 3 Cl₂(g) + 3 C(s) → 3 CO(g) + FeCl₂(s) + TiCl₄(g) TiCl₄(g) + 2 Mg(s) → 2 MgCl₂(l) + Ti(s) Suppose that an ilmenite–sand mixture contains 22.8% ilmenite by mass and that the first reaction is carried out with a 90.8% yield. If the second reaction is carried out with an 85.9% yield, what mass of titanium can be obtained from 1.00 kg of the ilmenite– sand mixture?

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Hi everyone here we have a question telling us that hematite is a common iron oxide compound with the formula iron three oxide and is widely found in rocks and soils. The iron can then be distracted from the hematite by the following set of reactions. Three, iron to iron three oxide plus carbon monoxide forms to iron to three oxide plus carbon dioxide, iron to three oxide plus carbon monoxide forms +33, iron to oxide plus carbon dioxide, iron to oxide plus carbon monoxide forms iron plus carbon dioxide. And it tells us to suppose that the 1st and 2nd reaction is carried out with an 85.6% yield. If the third reaction is carried out with a 95.6% yield, what massive iron can be obtained from 5.00 kg of hematite. So first we're going to calculate the molar mass Of Iron three Oxide. So we have iron Which has a molar mass of 55 .85, which can be found on the periodic table, times two Equals 111. And then we have oxygen, which is 16 times three, Which is 48 For a total of 159. g per mole. And now we need to calculate the moles of Iron three oxide. So we know we're starting with 5. kg And we want to change that to Graham. So we're going to multiply by 10 to the 3rd g over one kg And then multiply by one mole of Iron three oxide Over its smaller mass. So 159.7 g Of Iron three Oxide. And that equals 31. moles of iron three oxide. And now we're gonna calculate The amount of iron to three oxide from our first reaction. So we have three iron, three oxide plus carbon monoxide forms to iron to three oxide plus carbon dioxide. So we have 31. moles of Iron three oxide times two moles of our end to three oxide over three moles Of Iron, three oxide. And our moles of Iron three oxide are going to cancel out and that is going to equal 0.9 moles of iron to three oxide. And here are kilograms are canceling out and our grams are canceling out and our moles and we're left with moles. So now we want to consider the yield percent of 85.6%. So we have 0.9 moles of iron to three oxide Times 0.856 because we're dividing 85.6 by And that equals 17.9 moles of iron to three oxide. And now we're gonna calculate our moles of iron to oxide from our second reaction. So we have 17 0.9 moles of iron to three oxide, times three moles of iron to oxide. And that is the multiple ratio from our balanced equation over one mole of iron to three oxide. And that equals 53.7 moles of iron to oxide. And we're going to do the same thing we did for our and 23 oxide with the percent yield. So we're gonna take our 53.7 moles Of Iron to Oxide Times 0.856. And that equals 46.0 moles of iron, two oxide. And now we have to calculate our iron from our third reaction. So we have 46 0. moles of iron to oxide times one mole of iron. This is our multiple ratio over one mole of iron to oxide And that equals 46.0 moles of iron. And now it has a yield of 95.6%. So divided by 100 that's 0.956%. So 46.0 Moles of Iron Times 0. Equals 44.0 moles of iron. And now we're going to calculate our mass of iron. So the molar mass of iron, which is found on the periodic table is 0.85 grams per mole. So we're gonna take 44.0 moles of iron Times 55. g Permal. And our molds will cancel out giving us 2400 0.4 g. And we want to change that to kilograms. So we're gonna multiply by one kg Over 10 to the 3rd g. and our g are going to cancel out giving us 2.46 kilograms of iron. And that is our final answer. Thank you for watching. Bye.

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Key Concepts

Stoichiometry

Percent Yield

Mass Composition